State the parameters given in the question
[tex]n=450[/tex]The sample size n is 450
Hence, the first box n=450
[tex]X=55[/tex]The number of students X that attended private school is 55
Hence, the second box X=55
Therefore, sample proportion is
[tex]\begin{gathered} ^{\prime}p=\frac{X}{n}=\frac{55}{450} \\ ^{\prime}p=0.122 \end{gathered}[/tex]Hence, the third box is 'p=0.122
[tex]\begin{gathered} ^{\prime}q=1-^{\prime}p \\ ^{\prime}q=1-0.122 \\ ^{\prime}q=0.878 \end{gathered}[/tex]Hence, the forth box is 'q= 0.878
From the question, the confidence interval is 95% can be represented as
[tex]\begin{gathered} z_{\alpha I2}=95\text{ \%=0.95} \\ \alpha=1-0.95=0.05 \\ \alpha I2=\frac{0.05}{2}=0.025 \end{gathered}[/tex]The Z score for confidence interval of 95% is gotten from the table as 1.96
The formula for the confidence interval for p is
[tex]^{\prime}p-z_{\alpha I2}\sqrt[]{\frac{^{\prime}p^{\prime}q}{n}}Substitute into the formula
[tex]\begin{gathered} ^{\prime}p-z_{\alpha I2}z_{\alpha I2}\sqrt[]{\frac{^{\prime}p^{\prime}q}{n}} \\ ^{\prime}p=0.122;^{\prime}q=0.878;z_{\alpha I2}=1.96 \\ =0.122-1.96_{}\sqrt[]{\frac{0.122\times0.878}{450}} \\ =0.122-1.96\sqrt[]{\frac{0.107116}{450}} \\ =0.122-1.96\sqrt[]{0.0002380356} \\ =0.122-1.966(0.015428) \\ =0.122-0.03024 \\ =0.09176 \\ =0.092(3\text{ decimal places)} \end{gathered}[/tex][tex]\begin{gathered} z_{\alpha I2}\sqrt[]{\frac{^{\prime}p^{\prime}q}{n}}\text{was calculated to be 0.03024} \\ \text{Therefore} \\ ^{\prime}p+z_{\alpha I2}\sqrt[]{\frac{^{\prime}p^{\prime}q}{n}}=0.122+0.03024=0.15224 \\ =0.152(3\text{decimal places)} \end{gathered}[/tex]Hence, the confidence interval is 0.092
